Practically always occurs that a particular element like load is always variable when the other elements are fixed in a circuit. If we consider our household as example a household outlet terminal may be connected to different appliances constituting a variable load. Every time the variable is changed, the complete circuit has to be analyzed again. This is harassment for electrical engineer. To avoid this problem we can use the method of Thevenin’s theorem. The fixed circuit is replaced by an equivalent circuit and according to this method.

Figure 1(a) shows a linear circuit connected with a variable load, the load can be single resistor or different circuit.

According to Thevenin’s theorem we can replace figure 1(a) linear circuit with figure 1(b). Left side of the figure 1(b) is said Thevenin’s equivalent circuit. French telegraph engineer M.Leon Thevenin developed it in 1883.

Thevenin’s theorem states that ** “ a linear two terminal circuit can be replaced by an equivalent circuit which have a voltage source V_{Th} in series with a resistor R_{Th} where V_{Th} is the open circuit voltage at the terminals and R_{Th} is the input or equivalent resistance at the terminals when the independent sources are turned off””**.

We have to find the Thevenin equivalent voltage V_{Th} and resistance R_{Th. }For doing this let us consider figure 1(a) & figure 1(b) are equivalent. The two circuits will be equal if they have same voltage- current relationship at their two terminals.

Let us find that what will make two circuits are equivalent. If we remove the load from figure 1(a), the voltage source will be open circuit then voltage source of figure 1(a) must equal to voltage source V_{Th} in figure 1(b). So the two circuits will become equivalent and open circuit voltage across two terminals is

V_{Th} = V_{oc}

And figure 2(a) shows that

If the load is disconnected a-b terminals are open circuit. Now we turn off the all independent source the input equivalent resistance of figure 1(a) is equal to input resistance R_{Th} and figure 2(b) shows it and that is

R_{Th} = R_{in}

For applying this method we have to find out Thevenin’s resistance R_{Th} considering two cases.

**Case 1:**If the network has no dependent sources, we turn off all independent sources. R_{Th}is the input resistance of the network looking between terminals and b as shown in figure 2(b).

**Case 2:**If the network has dependent sources, we turn off all independent sources. As with superposition theorem, dependent sources are not to be turned off because they are controlled by circuit variables.

Applying a voltage source *v _{0}* at terminals a and b to determine resulting current

*i*.

_{0}

R_{Th} = *v _{0} / i_{0}*as shown in figure 3(a). We can apply a current source alternatively at a and b terminal as shown in figure 3(b).

The two processes will give same result. In either process we suppose any value of *v*_{0} and *i _{0}*. As example we can use v

_{0}= 1V or i

_{0}= 1A or even can use unspecified values of v

_{0}or i

_{0}.

The value of R_{Th} takes negative always. In this situation the negative resistance (v = – iR) values of R_{Th} tells that the circuit is supplying power.

**Importance of Thevenin’s theorem:**

In circuit analysis Thevenin’s theorem is very important. It is helpful simplifying a circuit. A large circuit can be replaced with an independent voltage source and a resistor. This replacement system is very helpful tool in circuit design.

According to Thevenin’s theorem we replace a huge circuit with Thevenin’s equivalent, actually original circuit behaves same as Thevenin’s equivalent.

Consider a linear two terminal circuit shown in figure 4(a).

Load current I_{L} and voltage across the load v_{L }can be easily determined if the Thevenin’s equivalent circuit at the load terminals is obtained shown in figure 4(b).

From figure 4(b) we get

Above equation shows Thevenin’s equivalent is a simple voltage divider, yielding V_{L} by mere inspection.

## Thevenin’s theorem problem**:**

**Find the Thevenin’s equivalent circuit of the circuit shown in figure 5, to the left of the terminals a-b. Then find the current through R _{L} = 6, 16, 36Ω.**

**Answer: **Turning off voltage source and making short circuit and 2A current circuit replacing it open circuit we get the circuit shown in figure 6 (a)

To find V_{Th} applying mesh analysis consider figure 6(b).

In two loops applying mesh analysis we get,

-32 + 4i_{1} + 12 (i_{1} – i_{2}) = 0, i_{2} = -2A

i_{1} = 0.5A.

In this way V_{Th} = 12 (i_{1} – i_{2}) = 30v

We can use alternative way nodal analysis or source transformation to find V_{Th}.

The Thevenin’s equivalent circuit shown in figure 7.

We know

I_{L} = V_{Th} / (R_{Th} + R_{L}) = 30 / (4 + R_{L})

When R_{L} = 6, I_{L} = 30/10 = 3A

When R_{L} = 16, I_{L} = 30/20 = 1.5A

When R_{L} = 36, I_{L} = 30/40 = 0.75A